Question: Putri is searching for a secret cave that she knows is $3.72$ kilometers from her home. Starting from her home, she walks $2.60$ kilometers in a straight line before coming upon a large rock. Climbing atop the rock, Putri spots the cave. She then turns $58^\circ$ to the right of her original path and walks to the cave. After exploring the cave, she looks back at the big rock and wonders how to get home. Assuming Putri is facing the big rock, how many degrees to her left should she turn before walking home? Do not round during your calculations. Round your final answer to the nearest degree.
Converting the problem into geometrical terms Our problem can be modeled by the following triangle $\triangle ABC$, where we want to find $\angle C=\theta$. $\angle B$ is supplementary to $58^\circ$, so $\angle B=122^\circ$. $2.60\text{ km}\,\,\,\,\,\,\,\,\,$ $\,\,\,\,\,\,\,\,\,\,3.72\text{ km}$ $A$ $B$ $C$ $\theta$ $\;\;122^\circ$ $58^\circ$ Since we are given two side lengths, we can use the law of sines. When using the law of sines we have to keep in mind the ambiguous case, where the angle can be either acute or obtuse. In our case, as the triangle already has an angle of $122^\circ$, we know $\theta$ must be acute so the ambiguous case doesn't apply. Using the law of sines $\begin{aligned} \dfrac{\sin(C)}{AB}&=\dfrac{\sin(B)}{AC}\\\\ \dfrac{\sin(\theta)}{2.6} &= \dfrac{\sin(122^\circ)}{3.72} \gray{\text{Substitute}} \\\\ \sin(\theta) &= \dfrac{2.6 \cdot \sin(122^\circ) }{3.72} \\\\ \theta &= \sin^{-1}\left(\dfrac{2.6 \cdot \sin(122^\circ) }{3.72}\right) \\\\ \theta &\approx 36^\circ \end{aligned}$ The answer Putri should turn $36^\circ$ to her left.